integral from 0 to 4 of (x^2)/(x^3+6)

Lösung

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 6} d x

\frac{1}{3} ln (\frac{35}{3})

Dezimale

0.81891 \dots

Schritte zur Lösung

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 6} d x

Wende U-Substitution an

= \int_{6}^{70} \frac{1}{3 u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{6}^{70} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{6}^{70}

Berechne die Grenzen:

ln (70) - ln (6)

= \frac{1}{3} (ln (70) - ln (6))

Vereinfache

= \frac{1}{3} ln (\frac{35}{3})

\int_{0}^{3} (\frac{1}{2} x^{3} - 3 x) d x

\int_{2}^{3} (4 + 2 y)^{2} d y

\int_{2}^{4} \frac{1}{x - 2} d x

\int_{- 3}^{- 2} x^{2} + 5 x + 6 d x

\int_{0}^{\frac{π}{3}} (4 sec^{2} (x)) d x